Answer» Correct Answer - Option 2 : (p ∧ q) → (~p) ∨ q
p | q | ~q | p ∨ ~q | ~p | p ∧ ~q | T | T | F | T | F | F | T | F | T | T | F | T | F | T | F | F | T | F | F | F | T | T | T | F |
p | q | p ∨ q | p → p ∨ q | p ∧ q | p ∧ q → p | T | T | T | T | T | T | T | F | T | T | F | T | F | T | T | T | F | T | F | F | F | T | F | T |
p | q | ~p ∨ q | (p ∧ q) → (~p) ∨ q | (p ∨ q) → (p ∨ (~q)) | T | T | T | T | T | T | F | F | T | T | F | T | T | T | F | F | F | T | T | T |
A tautology is a formula or assertion that is true in every possible interpretation. So, by the truth table (p ∨ q) → (p ∨ (~q)) statement is not a tautology. Atautologyis a statement that is alwaystrue, no matter what. If you construct atruth tablefor a statement and all of the column values for the statement aretrue(T), then the statement is a tautologybecause it's alwaystrue!
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